Question

You are evaluating 2 machines the investment of 2 mutually exclusive machines. Each machine has an eight year life and you plan to keep whichever machine you pick for the full 8 years. The firm's MARR is 10%. The cash flows for each machine are summarized in the following table:

A | B | |

Initial Cost | $4000 | $3000 |

Annual Benefit | $800 | $600 |

Annual Cost | $100 | $50 |

Salvage Value | $1500 | $1000 |

Each investment has an IRR greater than the 10% MARR. Using Incremental Replacement Analysis, which investment should be chosen and why (show calculations in answer or in written submission)?

Answer #1

t = 8 yrs

MARR = 10%

Base alternative will be B, as it has lower cost

Incremental initial cost (A-B) = 4000 - 3000 = 1000

Incremental annual cost (A-B) = 100 - 50 = 50

Incremental annual benefit (A-B) = 800 - 600 = 200

Incremental salvage value (A-B) = 1500 - 1000 = 500

Let incremental salvage value be i%, then

(200-50)*(P/A,i%,8) + 500*(P/F,i%,8) = 1000

150*(P/A,i%,8) + 500*(P/F,i%,8) = 1000

Dividing by 50

3*(P/A,i%,8) + 10*(P/F,i%,8) = 20

using trail and error method

When i = 10%, value of 3*(P/A,i%,8) + 10*(P/F,i%,8) = 3*5.334926 + 10*0.466507 = 20.669852

When i = 11%, value of 3*(P/A,i%,8) + 10*(P/F,i%,8) = 3*5.146123 + 10*0.433926 = 19.777633

using interpolation

i = 10% + (20.669852-20)/(20.669852-19.777633)*(11%-10%)

i = 10% + 0.75% = **10.75% (Approx)**

**As incremental IRR > MARR, option A should be
selected**

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